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Tanzania [10]
2 years ago
15

A car generator turns at 400 rpm (revolutions per minute) when the engine is idling. It has a rectangular coil with 300 turns of

dimensions 5.00 cm by 5.22 cm that rotates in an adjustable magnetic field. What is the field strength needed to produce a 24.0 V peak emf
Physics
1 answer:
Sav [38]2 years ago
6 0

The field strength needed to produce a 24.0 V peak emf is 0.73T.

To find the answer, we need to know about the expression of emf.

What's the expression of peak emf produced in a rotating rectangular loops?

  • The peak emf produced in a rotating loops= N×B×A×w
  • N= no. of turns of the loop, B= magnetic field, A= area of loop and w= angular frequency
  • So, B = emf/(N×A×w)
<h3>What's the magnetic field applied to the loop, when rectangular coil with 300 turns of dimensions 5.00 cm by 5.22 cm rotates at 400 rpm produce a 24.0 V peak emf?</h3>
  • N= 300, A= 5cm × 5.22cm = 0.05m × 0.0522m = 0.00261 m²
  • Emf= 24V, w= 2π×400 rpm= 2π×(400rps/60) = 42 rad/s
  • Now, B= 24/(300×0.00261×42)

B= 24/(300×0.00261×42) = 0.73T

Thus, we can conclude that the magnetic field is 0.73T.

Learn more about the electromagnetic force here:

brainly.com/question/13745767

#SPJ4

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newton's second law states that when a net force acts on an object, it accelerates it.Explain how it would be possible for two o
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Answer:

See Explanation

Explanation:

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A 15.0 Ohms resistor is connected in series to a 120V generator and two 10.0 Ohms resistors that are connected in parallel to ea
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Hi there! :)

Reference the diagram below for clarification.

1.

We must begin by knowing the following rules for resistors in series and parallel.

In series:
R_T = R_1 + R_2 + ... + R_n

In parallel:
\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}

We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.

\frac{1}{R_{T, parallel}} = \frac{1}{10} + \frac{1}{10}\\\\\frac{1}{R_{T, parallel}} = \frac{2}{10} = \frac{1}{5}\\\\R_{T, parallel} = 5\Omega

Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
R_T = 15 + 5\\\\\boxed{R_T = 20 \Omega}

2.

We can use Ohm's law to solve for the current in the circuit.

i = \frac{V}{R_T}\\\\i = \frac{120}{20} = \boxed{6 A}

3.

For resistors in series, both resistors receive the SAME current.

Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.

In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.

4.

Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.

V = iR\\\\V = (6)(15) = \boxed{90 V}

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